stokes theorem integral around a sphere and plane

Question

Evaluate using Stokes theorem $$\int_\gamma ydx+zdy+xdz$$ where $\gamma$ is the curve given by $(x-a)^2+(y-a)^2=z^2=2a^2$,$x+y=2a~~$, starting from $(2a,0,0)$ and then going below the z-plane. $$$$ What i tried was i calculate $curl(\vec{V})=-(i+j+k)~~$. Then i calculated the unit norma as $\hat{n}=\dfrac{(x-a)\hat{i}+(y-a)\hat{j}+z\hat{k}}{\sqrt{2}}$. Now $$\int\int_S(\nabla\times \vec{V})\cdot \hat{n}=\int\int_S(2a-x-y-z)dS=\int\int_R \sqrt{(x-a)^2+(y-a)^2}dxdy$$ where $R $ is the region bounded by the circle $(x-a)^2+(y-a)^2=2a^2$ $$$$ is this correct, if not please help. The problem with me using Stokes, Gauss theorems is that i am not able to handle the surfaces of integration, limits of integration. Please also suggest some books to make my understanding more clear

Answer 1

Since the curve $\gamma$ lays in the plane $x+y=2a$, why don't you consider $S$ as the intersection disc (not the sphere cap) and take the unit norm $\hat{n}=\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}$ which is constant? In this way $$(\nabla\times \vec{V})\cdot \hat{n}=-(\hat{i}+\hat{j}+\hat{k})\cdot \dfrac{\hat{i}+\hat{j}}{\sqrt{2}}=-\sqrt{2}.$$

Moreover, $\gamma$ is a great circle of the sphere $(x-a)^2+(y-a)^2+z^2=2a^2$, because the centre $(a,a,0)$ belongs to the plane $x+y=2a$. Hence the disk $S$ such that $\partial S=\gamma$ has area $2\pi a^2 $. Finally, by Stokes' Theorem $$\int_\gamma ydx+zdy+xdz=\iint_S (\nabla\times \vec{V})\cdot \hat{n}dS=-\sqrt{2}|S|=-2\sqrt{2} \pi a^2.$$ Note that $\hat{n}$ induces the correct orientation on $\gamma$: starting from $(2a,0,0)$ the path goes below the $z$-plane.